19
$\begingroup$

Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that: $$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3d}}+\frac{d}{\sqrt{d+3a}}\geq\sqrt{a+b+c+d}$$ I tried Holder, AM-GM and more, but without success.

$\endgroup$
  • $\begingroup$ I have an advice, set $a+3b=s^2,b+3c=t^2.\cdots$ $\endgroup$ – Isaac YIU Math Studio 2 days ago
  • $\begingroup$ @Isaac YIU Math Studio Let $a+3b=x^2$, $b+3c=y^2$, $c+3d=z^2$ and $d+3a=t^2$, where $x$, $y$, $z$ and $t$ are positives. Thus, $a=\frac{-x^2+3y^2-9z^2+27t^2}{80}$ and it's nothing. $\endgroup$ – Michael Rozenberg 2 days ago
  • $\begingroup$ I think it may have help $\endgroup$ – Isaac YIU Math Studio 2 days ago
  • $\begingroup$ I don't see it. :( $\endgroup$ – Michael Rozenberg 2 days ago
  • $\begingroup$ Not nothing! It made a better expansion of the left side $\endgroup$ – Isaac YIU Math Studio 2 days ago
1
$\begingroup$

Hello Michael Rozenberg . I start my proof with a classical substitution .

Put: $A=a$; $AB=b$; $AC=c$; $AD=d$ We get : $$\frac{1}{\sqrt{1+3B}}+\frac{B}{\sqrt{B+3C}}+\frac{C}{\sqrt{C+3D}}+\frac{D}{\sqrt{D+3}}\geq \sqrt{1+B+C+D}$$

Make a new substitution like this :

$x=B$; $C=xh$; $D=x\beta$

We get : $$\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\frac{x\beta}{\sqrt{x\beta+3}}\geq \sqrt{1+x(1+h+\beta)}$$ We can rewriting the inequality like this : $$\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\sqrt{x\beta+3}+\frac{-3}{\sqrt{x\beta+3}}-\sqrt{1+x(1+h+\beta)}\geq 0$$

Now we introduce the following function :

$$f(x)=\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\sqrt{x\beta+3}+\frac{-3}{\sqrt{x\beta+3}}-\sqrt{1+x(1+h+\beta)}$$

The idea is to integrate the function $f$ we get :

$$F(x)=\frac{2}{3}\sqrt{1+3x}+\frac{2}{3}(x)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}}+\frac{2}{3\beta}(x\beta+3)^{\frac{3}{2}}-\frac{6}{\beta}\sqrt{x\beta+3}-\frac{2}{3(1+h+\beta)}(1+x(1+h+\beta))^{\frac{3}{2}}$$

Now the idea is to note that the function $f$ is positiv if the function $F$ is increasing . Namely when we have : $F(X)\leq F(Y)$ with $X\leq Y$

Here we take $X=x-\epsilon$ and $Y=y+\epsilon$

We cut the function $F$ in others to a question of space

We have :

$g(x-\epsilon)-g(x+\epsilon)=\frac{2}{3}\sqrt{1+3(x-\epsilon)}-\frac{2}{3}\sqrt{1+3(x+\epsilon)}$

$u(x-\epsilon)-u(x+\epsilon)=\frac{2}{3}(x-\epsilon)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x-\epsilon)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}}-(\frac{2}{3}(x+\epsilon)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x+\epsilon)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}})$

$v(x-\epsilon)-v(x+\epsilon)=\frac{2}{3\beta}((x-\epsilon)\beta+3)^{\frac{3}{2}}-\frac{2}{3\beta}((x+\epsilon)\beta+3)^{\frac{3}{2}}$

$q(x+\epsilon)-q(x-\epsilon)=\frac{6}{\beta}\sqrt{(x+\epsilon)\beta+3}+\frac{2}{3(1+h+\beta)}(1+(x+\epsilon)(1+h+\beta))^{\frac{3}{2}} -(\frac{6}{\beta}\sqrt{(x-\epsilon)\beta+3}+\frac{2}{3(1+h+\beta)}(1+(x-\epsilon)(1+h+\beta))^{\frac{3}{2}} )$

Now we have the following elementary inequalities with $n\geq x^6$:

$g(x-\epsilon)-g(x+\epsilon)\leq -\frac{2}{3}\frac{\epsilon}{n} x^6$

$u(x-\epsilon)-u(x+\epsilon)\leq -\frac{2\epsilon}{3n}x^6(\frac{1}{\sqrt{1+3h}}+\frac{h}{\sqrt{h+3\beta}})$

$v(x-\epsilon)-v(x+\epsilon)\leq -\frac{2\epsilon}{3n}x^6$

$q(x+\epsilon)-q(x-\epsilon)\leq \frac{\epsilon}{n}(\frac{2}{3})x^6$

So you have just to add the elementary inequalities to get the result (and derivate) If you have questions tell me .

$\endgroup$
  • $\begingroup$ I think there are mistakes in your integration. $\endgroup$ – Michael Rozenberg Mar 26 '17 at 18:25
  • $\begingroup$ I am going to correct .Thanks! $\endgroup$ – max8128 Mar 27 '17 at 8:21
1
$\begingroup$

WLOG, assume that $d = \min(a,b,c,d)$.

Squaring both sides and using AM-GM, it suffices to prove that \begin{align} &\sum_{\mathrm{cyc}} \frac{a^2}{a+3b} + \sum_{\mathrm{cyc}} \frac{4ab}{a+3b + b+3c} + \frac{4ac}{a+3b + c+3d} + \frac{4bd}{b+3c + d+3a} \\ \ge\ & a + b + c + d. \end{align} After clearing the denominators, it suffices to prove that $f(a, b, c, d)\ge 0$ where $f(a,b,c,d)$ is a homogeneous polynomial of degree $11$.

The Buffalo Way works. Let $c=d+s, \ b = d+t, \ a = d+r; \ s, t, r\ge 0$. We have $$f(d+r, d+t, d+s, d) = q_9d^9 + q_8d^8 + \cdots + q_1d + q_0.$$ Here $q_9 = 12582912r^2+16777216rs-20971520rt+12582912s^2-20971520st+12582912t^2$, etc.

It suffices to prove that $q_9, q_8, \cdots, q_0 \ge 0$. I used Mathematica Resolve and the Buffalo Way to verify it. We are done. However, nice or simple proof for $q_9, q_8, \cdots, q_0 \ge 0$ is expected.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.